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Chemistry: Solving Equilibrium Problems


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About this Lesson

  • Type: Video Tutorial
  • Length: 10:42
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 115 MB
  • Posted: 07/14/2009

This lesson is part of the following series:

Chemistry: Full Course (303 lessons, $198.00)
Chemistry: Final Exam Test Prep and Review (49 lessons, $64.35)
Chemistry: Chemical Equilibrium (14 lessons, $20.79)
Chemistry: Using Equilibrium Constants (5 lessons, $8.91)

This lesson was selected from a broader, comprehensive course, Chemistry, taught by Professor Harman, Professor Yee, and Professor Sammakia. This course and others are available from Thinkwell, Inc. The full course can be found at The full course covers atoms, molecules and ions, stoichiometry, reactions in aqueous solutions, gases, thermochemistry, Modern Atomic Theory, electron configurations, periodicity, chemical bonding, molecular geometry, bonding theory, oxidation-reduction reactions, condensed phases, solution properties, kinetics, acids and bases, organic reactions, thermodynamics, nuclear chemistry, metals, nonmetals, biochemistry, organic chemistry, and more.

Dean Harman is a professor of chemistry at the University of Virginia, where he has been honored with several teaching awards. He heads Harman Research Group, which specializes in the novel organic transformations made possible by electron-rich metal centers such as Os(II), RE(I), AND W(0). He holds a Ph.D. from Stanford University.

Gordon Yee is an associate professor of chemistry at Virginia Tech in Blacksburg, VA. He received his Ph.D. from Stanford University and completed postdoctoral work at DuPont. A widely published author, Professor Yee studies molecule-based magnetism.

Tarek Sammakia is a Professor of Chemistry at the University of Colorado at Boulder where he teaches organic chemistry to undergraduate and graduate students. He received his Ph.D. from Yale University and carried out postdoctoral research at Harvard University. He has received several national awards for his work in synthetic and mechanistic organic chemistry.

About this Author

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We understand that a chemical reaction can be described by the equilibrium constant, a number that tells us what side of the reaction is favored--a balance point, if you will, that tells us whether the reactants are in the majority or the products are in the majority once we reach equilibrium. We also have introduced the idea of an equilibrium quotient, which is just simply the ratio of the products and the reactants, and by comparing that ratio to the equilibrium constant, which is a ratio, we can evaluate which direction the reaction needs to go to get to equilibrium.
Well, let's look at another example. This is methane plus water goes to carbon monoxide and hydrogen. In this reaction, inspection of the equilibrium constant, K[p]--and now notice I'm defining my equilibrium this time in terms of partial pressures--my K is very small, 1.8x10^-7. That tells me that at equilibrium, this reaction will be heavily favoring this side, meaning that at equilibrium I'll have primarily methane and water, and very small amounts of these materials unless I have an excess of this guy to begin with. But again, the equilibrium will be shifted far to this side.
However, let's go ahead a look at a set of initial conditions and see what that says about which way the reaction is going to move. Let's suppose that I started with 1.4 atmospheres of methane, 2.3 atmospheres of water, 1.6 atmospheres of carbon monoxide, and no hydrogen. Although the equilibrium tells me that equilibrium will be heavily favoring this side, notice that I have no hydrogen to begin with. I can calculate Q, the reaction quotient for this equilibrium system, and let's go ahead and do that. Again, Q is partial pressure--it's going to look exactly like the equilibrium constant here. If we plug in numbers for those partial pressures, this zero in the numerator is going to make Q zero. So we have a situation where Q is less than K, even though K is very, very small. And so what that means is that although the equilibrium will favor this side, the reaction where we're starting it has to move from left to right to establish the equilibrium. It's just that it's not going to be moving very far. It's only going to be moving to the point where it generates a little bit of hydrogen and then equilibrium will be established. So again, key points: the equilibrium here is going to favor the left side, but we're starting with no hydrogen so the reaction has to move in a forward direction to establish the equilibrium.
Now an interesting question would be to ask, "What will be the partial pressure of hydrogen once we reach equilibrium?" So that's what we're after here. How do we solve such a thing? Okay, well, what we use--chemists often use, especially when you're first being introduced to equilibrium problems, something called an ICE diagram. So let me describe how that looks for you. What I'm going to do first is simply list the different components of my reaction up at the top here. And I find it convenient to list them in the order of the reaction. So here then down below is I, standing for "initial concentrations. " We already know those initial partial pressures in this case are 1.4, 2.3, and 1.6 atmospheres.
Now, what I'm really interested in is, what is the change going from my initial conditions to my equilibrium? How much does this reaction have to go in the forward direction to reach that equilibrium value? So my next entry is going to be a C, corresponding to "change." So here I'm going to describe my change. I don't know how much it's going to change. I know my equilibrium constant but I don't know, just from inspection, how much it's going to shift. But I'll identify that as a variable, X, so I know that my methane--because my Q is smaller than K, I know the reaction has to go this way--so the amount of methane I'm going to have is going to decrease by a value, I'll say X. So we'll call this then -X, that's the change in methane. Likewise, the amount of water will change by the same amount. Meanwhile, I will gain that much carbon monoxide and I will gain three times that much of the hydrogen--because notice my ratio here, 1-to-3. For every one methane I consume, I gain three equivalents of the hydrogen.
Finally, in order to relate this to the equilibrium expression, I need to know my equilibrium partial pressures. Remember this is only valid at equilibrium, so I need to know my partial pressure at equilibrium. I don't know what it is but I can describe it in terms of my initial concentration and my change in concentration. Now, I use "partial pressure" and "concentration" interchangeably sometimes, and really I should be a little more careful about that. They are both effectively concentrations but the difference is, of course, the units are different. So I'll try to be a little more careful. I'm talking rigorously here in terms of partial pressures not in terms of concentrations.
So what are my equilibrium partial pressures? 1.4 atmospheres minus the amount we lost; 2.3 atmospheres minus the amount we lost; 1.6 atmospheres plus the amount that we gained; and three times that much. So this would be, then, described in terms of the variable X, my final equilibrium concentration. So now I'm set to take these numbers and plug them into this expression where I know their relationship at equilibrium.
Okay, here we go. So in fact, let's take this back up here so I can keep track of the order. All right. I'm first interested in our partial pressure of carbon monoxide. That's going to be 1.6 atmospheres plus the amount that we gained. A partial pressure of hydrogen is going to be 3X, and remember this is cubed because we have that coefficient of 3. Divide by the partial pressure of methane, that's 1.4 minus the amount we lost. And then the partial pressure of water was 2.3 atmospheres minus the amount that we lost. And that has a value of 1.8x10^-7 at this temperature. Okay?
Now, at this stage all we need to do is evaluate for X and we've solved it. Well, all you need to do is go get yourself the Thinkwell version of algebra to solve this mess because there is no way we're going to be solving it right here. We either go and get a graphics calculator or do something. We're going to have a fourth degree polynomial here. Let's try finding some type of a simplifying assumption that will make our lives easier.
Think about the original equilibrium. We know that this equilibrium, when established, is going to lie heavily to this side. We've already established that because K is so small, we're going to be making very little hydrogen. In other words, X, the amount of change, is going to be very small because we're very close to equilibrium already. Let's try the assumption--nothing to lose here--of X is equal to a number so small that it's negligible compared to 1.6 or 1.4 or 2.3. Notice these are simple additions or subtractions. So a small number compared to a big number, the big number is going to be what dominates the answer. We don't want to make any approximations here, though. This is not added or subtracted from anything. So what we'll write down is 1.6 atmospheres times 3X^3, divided by 1.4 atmospheres and 2.3 atmospheres, with again our crucial assumption being that X is small compared to any of these big numbers here. That still is equal to 1.8x10^-7. So we end up with X^3 then, if we work this out, equal to 1.3x10^-8, or X is equal, in this case, to 2.4x10^-3, and that's again in atmospheres.
Now remember, X is in fact not the partial pressure of hydrogen. X was the change in the equilibrium. So specifically, it was the amount of methane lost or the amount of water lost. The actual partial pressure of hydrogen is in fact going to be three times X, and so let's write that down then. 3X is equal to the partial pressure of hydrogen, and that has a value of 7.1x10^-3 atmospheres. And so indeed that partial pressure is small. And again I urge you to use your intuition and common sense as much as you can here. We know from K that that equilibrium was small, so the amount of hydrogen that we are generating to reach equilibrium should indeed be a very small value compared to the amount of things that we started with.
So what have we done? Well, we've developed something called an ICE diagram, standing for initial concentrations or partial pressures, change in concentrations, and the equilibrium values for concentrations or partial pressures. And by being able to express what those equilibrium concentrations are at equilibrium, we can relate it to the equilibrium constant and then solve ultimately for what the concentrations or what the partial pressures are for various components in our reaction.
What we're now going to do is look at a different type of problem that you might see where, unlike what we saw here, we start out far away from equilibrium rather than very close to equilibrium, and we'll see what strategy we need to use in that situation.
Chemical Equilibrium
Using Equilibrium Constants
Strategies for Solving Equilibrium Problems Page [2 of 2]

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